Frobenius linear systems have nonreduced general members

On the overlap $W_{01}=U_0\cap U_{\mathrm{\infty }}$ one has $u=t^{-1}$ and $e_{\mathrm{\infty }}=t^p{e}_0$ by [F3]. The pairs of functions $(1,u^p)$ and $(t^p,1)$ satisfy the compatibility $1=t^p{u}^p$ and $t^p=t^p\cdot 1$, so they define global sections $X^p,Y^p\in \mathrm{\Gamma }(X,L)$; their restrictions to $U_0$ are the polynomials $1$ and $t^p$, which are $k$-linearly independent, so $W={\mathrm{span}}_k\{X^p,Y^p\}$ is a $2$-dimensional subspace and, by [F1], a linear system on $X$ with $L=\mathcal{O}(p)$ invertible.

Every element of $k$ is a $p$th power: for $c\in k$ the polynomial $z^p-c$ is nonconstant, so it has a root $\alpha \in k$ by [F4], and ${\alpha }^p=c$. The characteristic $p$ is prime, hence $p\ge 2$, by [F16].

For a parameter $[a:b]\in \mathbf{P}(W)$ choose $\alpha ,\beta \in k$ with ${\alpha }^p=a$ and ${\beta }^p=b$ by step 1.2; then $(\alpha ,\beta )\mathrm{\ne }(0,0)$. By [F5] and [F6], $(z_1+z_2{)}^p=z_1^p+z_2^p$ in every commutative ring of characteristic $p$, so $a{X}^p+b{Y}^p=(\alpha X{)}^p+(\beta Y{)}^p=(\alpha X+\beta Y{)}^p$ in $k[X,Y]$. Hence the member $Z(a{X}^p+b{Y}^p)$ equals $Z((\alpha X+\beta Y{)}^p)$ as a closed subscheme of $X$, and the linear form $\alpha X+\beta Y$ is nonzero.

By [F3], $L=\mathcal{O}(p)$ is trivialized on the chart $U_0$ by $e_0$ and on $U_{\mathrm{\infty }}$ by $e_{\mathrm{\infty }}$, and on $U_0$ the section $a{X}^p+b{Y}^p$ restricts to $(a+b{t}^p)e_0$, while on $U_{\mathrm{\infty }}$ it restricts to $(a{u}^p+b)e_{\mathrm{\infty }}$. Therefore [F2] computes the member chart by chart: $Z(a{X}^p+b{Y}^p)\cap U_0=\mathrm{Spec}k[t]\mathrm{/}(a+b{t}^p)$ and $Z(a{X}^p+b{Y}^p)\cap U_{\mathrm{\infty }}=\mathrm{Spec}k[u]\mathrm{/}(a{u}^p+b)$, using the ideal generated by the local equation itself, with no reduction.

Local form of every member. If $\beta \mathrm{\ne }0$, then by [F5] and [F6] applied in $k[t]$, $a+b{t}^p=(\alpha +\beta t{)}^p={\beta }^p(t-\gamma {)}^p$ with $\gamma =-\alpha \mathrm{/}\beta$, so $Z(s)\cap U_0=\mathrm{Spec}k[t]\mathrm{/}((t-\gamma {)}^p)$ and the substitution $\tau =t-\gamma$ identifies it with $\mathrm{Spec}k[\tau ]\mathrm{/}({\tau }^p)$, a nonempty single point. If $\beta =0$, then $\alpha \mathrm{\ne }0$ and $a{u}^p+b=(\alpha u{)}^p$, so $Z(s)\cap U_{\mathrm{\infty }}=\mathrm{Spec}k[u]\mathrm{/}((\alpha u{)}^p)$ and the substitution $\tau =\alpha u$ identifies it with $\mathrm{Spec}k[\tau ]\mathrm{/}({\tau }^p)$. In both cases one chart of $Z(s)$ is $\mathrm{Spec}k[\tau ]\mathrm{/}({\tau }^p)$; the other chart of $Z(s)$ either is empty or, on the overlap, describes the same single point, and no chart adds a second point.

The system is base-point-free. By step 3.1 the member $Z(X^p)$, which is the parameter $[1:0]$, has its unique point in the chart $U_{\mathrm{\infty }}$ at $\tau =u=0$, namely $[0:1]$; the member $Z(Y^p)$, the parameter $[0:1]$, has its unique point in the chart $U_0$ at $\tau =t=0$, namely $[1:0]$. These two points of ${\mathbb{P}}_k^1$ are distinct, so $\mathrm{Bs}(W)\subseteq \mathrm{\mid }Z(X^p)\mathrm{\mid }\cap \mathrm{\mid }Z(Y^p)\mathrm{\mid }=\mathrm{\varnothing }$ and $\mathrm{Bs}(W)=\mathrm{\varnothing }$ by [F1].

Local ring and dimension at the point. By step 3.1 every member $Z(s)$ has local ring $k[\tau ]\mathrm{/}({\tau }^p)$ at its unique point $P$. In this ring every prime contains the nilpotent $\tau$ since ${\tau }^p=0$, and $k[\tau ]\mathrm{/}({\tau }^p)\mathrm{/}(\tau )\cong k$ is a field, so $(\tau )$ is the only prime ideal; hence $\dim k[\tau ]\mathrm{/}({\tau }^p)=0$ by [F10]. The ring has $k$-basis $1,\tau ,\dots ,{\tau }^{p-1}$, so it is finite dimensional over $k$ and Noetherian by [F11]; it is nonzero, and its $k$-dimension is $p$, the multiplicity of the fat point.

Tangent space of the point. By [F7] the tangent space $T_{P}Z(s)$ depends only on the local ring ${\mathcal{O}}_{Z(s),P}=k[\tau ]\mathrm{/}({\tau }^p)$, so it may be computed in the affine chart $\mathrm{Spec}k[\tau ]\mathrm{/}({\tau }^p)$ at its rational point $\tau =0$. The Jacobian matrix of the single equation ${\tau }^p$ is the $1\times 1$ matrix $(p{\tau }^{p-1})$, which is the zero matrix in characteristic $p$, so [F8] gives $T_{P}Z(s)\cong \ker (0)=k^1$ and ${\dim }_k{T}_{P}Z(s)=1$.

The point is not regular. The scheme $Z(s)$ is covered by the affine charts of steps 3.1 and 4.2, whose coordinate rings $k[\tau ]\mathrm{/}({\tau }^p)$ are Noetherian by step 4.2, so $Z(s)$ is locally Noetherian by [F12]. The field $k$ is algebraically closed, so the point $P$ has residue field $k$. By [F9], $P$ is regular exactly when ${\dim }_k{T}_{P}Z(s)=\dim {\mathcal{O}}_{Z(s),P}$; here that reads $1=0$, which is false, so $P$ is not a regular point and $Z(s)$ is not regular at $P$.

No member is smooth over $k$. The scheme $Z(s)$ is of finite type over $k$: it is covered by at most two affine charts $\mathrm{Spec}k[\tau ]\mathrm{/}({\tau }^p)$ and possibly an empty chart, each coordinate ring a quotient of $k[\tau ]$, hence of finite type over $k$ by [F15], so [F14] applied over the affine base $\mathrm{Spec}k$ makes $Z(s)\to \mathrm{Spec}k$ of finite type. Suppose it were smooth. Then by [F13], which carries AC, every local ring of the base change along $K=k$, that is of $Z(s)$ itself, would be regular, contradicting the nonregular local ring of step 6.1. Hence $Z(s)$ is not smooth over $k$.

Conclusion and scope. Every parameter $[s]\in \mathbf{P}(W)$ has a non-smooth member by step 7.1, while $\mathbf{P}(W)$ is nonempty and $W$ is base-point-free by step 4.1. So there is a nonempty Zariski-open subset of $\mathbf{P}(W)$ — namely all of it — on which every member is not smooth, and by [F1] the general member of $W$ is not smooth; the false Bertini claim for arbitrary base-point-free linear systems in characteristic $p$ is refuted by this smooth projective example $X={\mathbb{P}}_k^1$. The members are the $p$-fold points $(\alpha X+\beta Y{)}^p$ of step 2.1, of multiplicity $p\ge 2$ by steps 1.2 and 4.2, so the failure is the purely inseparable one and requires no hypothesis beyond algebraically closed characteristic $p$; the example uses $L=\mathcal{O}(p)$ rather than $\mathcal{O}(1)$, so the hyperplane case is not addressed.